CH.2. DEFORMATION AND STRAIN. Continuum Mechanics Course (MMC) - ETSECCPB - UPC

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1 CH.. DEFORMATION AND STRAIN Continuum Mechanics Course (MMC) - ETSECCPB - UPC

2 Overview Introduction Deformation Gradient Tensor Material Deformation Gradient Tensor Inverse (Spatial) Deformation Gradient Tensor Displacements Displacement Gradient Tensors Strain Tensors Green-Lagrange or Material Strain Tensor Euler-Almansi or Spatial Strain Tensor Variation of Distances Stretch Unit elongation Variation of Angles

3 Overview (cont d) Physical interpretation of the Strain Tensors Material Strain Tensor, E Spatial Strain Tensor, e Polar Decomposition Volume Variation Area Variation Volumetric Strain Infinitesimal Strain Infinitesimal Strain Theory Strain Tensors Stretch and Unit Elongation Physical Interpretation of Infinitesimal Strains Engineering Strains Variation of Angles 3

4 Overview (cont d) Infinitesimal Strain (cont d) Polar Decomposition Volumetric Strain Deformation Rate Spatial Velocity Gradient Tensor Strain Rate Tensor and Rotation Rate Tensor or Spin Tensor Physical Interpretation of the Tensors Material Derivatives Other Coordinate Systems Cylindrical Coordinates Spherical Coordinates 4

5 .1 Introduction Ch.. Deformation and Strain 5

6 Deformation Deformation: transformation of a body from a reference configuration to a current configuration. Focus on the relative movement of a given particle w.r.t. the particles in its neighbourhood (at differential level). It includes changes of size and shape. 6

7 . Deformation Gradient Tensors Ch.. Deformation and Strain 7

8 Continuous Medium in Movement Ω 0 : non-deformed (or reference) configuration, at reference time t 0. X : Position vector of a particle at reference time. Ω or Ω t : deformed (or present) configuration, at present time t. x : Position vector of the same particle at present time. X,t t 0 Q t Reference or non-deformed dx Ω t dx Q Ω 0 X P x P Present or deformed 8

9 Fundamental Equation of Deformation The Equations of Motion:,,,,,, 1,,3, t, t not x X X X t x X X X t i i i 1 3 i 1 3 not x X x X Differentiating w.r.t. X : x, not i X t dxi dx j Fij t dx j i j X j x X, t not dx dxfx, tdx X (material) deformation gradient tensor X,, 1,,3 Fundamental equation of deformation 9

10 Material Deformation Gradient Tensor 10 The (material) deformation gradient tensor: F(X,t): not FX, tx( X, t) not xi Fij i, j1,,3 X j x is a primary measure of deformation x x x X X X x x x x X1 X X3 X1 X X3 x 3 x3 x3 x 3 F x x T X X X 1 3 REMARK The material Nabla operator is defined as: ˆi X e characterizes the variation of relative placements in the neighbourhood of a material point (particle). dxf X, t dx i X 1 X X 3

11 Inverse (spatial) Deformation Gradient Tensor The inverse Equations of Motion: not 1 i i 1 3 i 1 3,,,,,, 1,,3 X x x x t X x x x t i not 1 X x X x, t, t Differentiating w.r.t. x : X X, t dx dx F t dx i j not i 1 i j ij j x j Xx, t X x F x x x not 1 d d, t d x,, 1,,3 Inverse (spatial) deformation gradient tensor 11

12 Inverse (spatial) Deformation Gradient Tensor The spatial (or inverse) deformation gradient tensor: 1 d, t d X F x x F -1 (x,t): 1 F x X x F 1 ij, t (, t) X x j i is a primary measure of deformation i, j 1,,3 X1 X1 X 1 T x1 x x3 X 1 1 X X X X F X x1 x x3 x1 x x3 X 3 X3 X3 X 3 X x1 x x3 REMARK The spatial Nabla operator is defined as: eˆi characterizes the variation of relative placements in the neighbourhood of a spatial point. It is not the spatial description of the material deformation gradient tensor x i x 1 x x3 1

13 Properties of the Deformation Gradients The spatial deformation gradient tensor is the inverse of the material deformation gradient tensor: x X x X x x i k i 1 1 ij k j j FF F F 1 If F is not dependent on the space coordinates, deformation is said to be homogeneous. FX (, t) Every part of the solid body deforms as the whole does. F t the The associated motion is called affine. x 1 If there is no motion, xx and F F 1. X 13

14 Example Compute the deformation gradient and inverse deformation gradient tensors for a motion equation with Cartesian components given by, X Ze Yt x Y1t 1 Using the results obtained, check that FF 1. t 14

15 Example - Solution X Ze Yt x Y1t t The Cartesian components of the deformation gradient tensor are, X Y t 1 Yt 0 T FX, t x x Y 1 t,, 0 1 t 0 t X Y Z t Ze 0 0 e The Cartesian components of the inverse motion equation will be given by, X x y Y 1 t t Z ze 1 X x, t yt 1 t yt 1 0 (1 t) FX( x, t), t fx, t 0 1 t 0 t fx, t 0 0 e 15

16 Example - Solution The Cartesian components of the inverse deformation gradient tensor are, 1 And it is verified that FF 1 : yt t 1 1 F x, t t t 0 0 e yt yt yt yt t 1 t 1 t (1 t) t 0 1 t t 1 t 1 t 0 0 e t t t 0 0 e 0 0 ee 1 F F 1 16

17 .3 Displacements Ch.. Deformation and Strain 17

18 Displacements Displacement: relative position of a particle, in its current (deformed) configuration at time t, with respect to its position in the initial (undeformed) configuration. Displacement field: displacement of all the particles in the continuous medium. Material description (Lagrangian form):, t, t X, X, 1,,3 U X x X X U t x t X i i i i Spatial description (Eulerian form): Ω 0, t, t x, x, 1,,3 u x x X x u t x X t i i i i t 0 X P U=u x Ω t P 18

19 Displacement Gradient Tensor, t, t X, X, 1,,3 U X x X X Ui t xi t Xi i Ui Jij Fij ij i, j1,,3 X j def JX, t UX, t F1, t, t x, x, 1,,3 u x x X x ui t xi Xi t i ui 1 jij ij Fij i, j1,,3 x j def 1 jx, t ux, t 1 F REMARK If motion is a pure shifting:. 19 Differentiating U w.r.t. X : def Ui( X, t) xi( X, t) Xi F J X j X j X j F ij Material Displacement Gradient Tensor Differentiating u w.r.t. x : Spatial Displacement Gradient Tensor ij ij ij ij def ui( x, t) xi Xi( x, t) 1 F xj xj xj F1 ij ij x X ij ij ij ij 1 xx (, t) X U( t) F 1 F and j J 0 j

20 .4 Strain Tensors Ch.. Deformation and Strain 0

21 Strain Tensors F characterizes changes of relative placements during motion but is not a suitable measure of deformation for engineering purposes: It is not null when no changes of distances and angles take place, e.g., in rigid-body motions. Strain is a normalized measure of deformation which characterizes the changes of distances and angles between particles. It reduces to zero when there is no change of distances and angles between particles. 1

22 Strain Tensors Consider F X,t Ω 0 t 0 X ds Q P dx x Ω t dx ds P Q dx F dx dx d F dx i ij j -1 X F dx dx F dx 1 i ij j where ds is the length of segment dx : ds dx dx and ds is the length of segment dx : ds dxdx

23 Strain Tensors d -1 XF dx dx F dx 1 i ij j dx F dx dx F dx i ij j ds dxdx ds d x d One can write: x T T T ds dxdxdx dxfdx FdXdXF FdX ds dx dx F dx F dx dx F F dx dx F F dx T k k ki i kj j i ki kj j i ik kj j not T 1 T 1 T 1 ds dxdxdx dx F dx F dx dxf F dx T 1 ds dx k dx k Fki dxifkj dxj dxifki Fkj dxj dxifik Fkj dxj REMARK The convention not 1 T T () () is used. 3

24 Green-Lagrange Strain Tensor T ds dxf FdX Subtracting: The Green-Lagrange or Material Strain Tensor is defined: 1 T EX, t F F1 1 Eij t Fki Fkj ij i j E is symmetrical: ds dx dx T T T 1 1 ds ds dxf FdXdXdXdXF FdXdX dxdx F F dx dxe dx def E 4 X,, 1,,3 T 1 T T 1 T T T T 1 T E F F1 F F 1 F F1E E E i, j1,,3 ij ji

25 Euler-Almansi Strain Tensor ds d d Subtracting: x x T 1 ds dxf F dx T 1 T 1 ds ds dxdxdxf F dxdx1dxdxf F dx 5 T 1 dx 1 F F dxdxedx def e The Euler-Almansi or Spatial Strain Tensor is defined: 1 T 1 ex, t 1 F F eij x, t ij Fki Fkj i, j 1,,3 e is symmetrical: T 1 T 1 T 1 T 1 T T T 1 T 1 e 1F F 1 F F 1F F e e e i, j1,,3 ij ji def e

26 Particularities of the Strain Tensors The Green-Lagrange and the Euler-Almansi Strain Tensors are different tensors. They are not the material and spatial descriptions of a same strain tensor. They are affected by different vectors (dx and dx) when measuring distances: ds ds dxe dxdxe dx The Green-Lagrange Strain Tensor is inherently obtained in material description, E EX,t. EE X x, t, t E x, t By substitution of the inverse Equations of Motion,. The Euler-Almansi Strain Tensor is inherently obtained in spatial description, e ex,t. ee x X, t, t e X, t By substitution of the Equations of Motion,. 6

27 Strain Tensors in terms of Displacements 1 Substituting F 1-j and FJ1 into 1 T 1 T 1 : 1 and 1 E F F e F F 1 T 1 T T E ( ) ( ) 1J 1J 1 JJ J J 1 U U i j Uk U k Eij i, j{1,, 3} X j Xi Xi X j 1 T 1 T T e 1 1j 1j jj j j 1 u u i j uk u k eij i, j1,,3 xj xi xi xj 7

28 Strain Tensors in terms of Displacements Proof: 1 T 1 T T T T E X, t J1 J1 1 J 1 JJ T T T T T T T T J J 1 J J J J J J 1 1 J J J J J T J 1 1 T 1 T T T T e 1 1j 1j 1 1 j 1 1 j j T T T T T T T T T j 1 1 j j j 1 1 j j j j j j j j 1 j j 8

29 Example For the movement in the previous example, obtain the strain tensors in the material and spatial description. X Ze Yt x Y1t t 9

30 Example - Solution The deformation gradient tensor and its inverse tensor have already been obtained: yt t The material strain tensor : t Yt F 0 1 t 0 F 0 0 t 1 t 0 0 e t 0 0 e Yt Yt 0 1 T 1 1 E F F1 Yt 1 t t 0 Yt Yt 1 t t t t t 0 0 e 0 0 e 0 0 ee 1 0 Yt 0 1 Yt Yt 1 t 1 0 e 30

31 Example - Solution The spatial strain tensor : t t 1 T 1 1 yt yt yt yt yt 11 0 e F F 1 t 1 t 1 t 1t 1t 1t t t e t t e e e yt t 1 yt yt t 1t 1 t t e 31

32 Example - Solution X Ze Y t x Y1t t In conclusion, the material strain tensor is: material description 0 Yt 0 1 EX, t Yt Yt 1t 1 0 t 0 0 e 1 And the spatial strain tensor is: Y y 1 t spatial description yt t 1 yt yt Ex, t 1t 1 0 1t 1t t 0 0 e 1 yt t 1 yt yt 1 ex, t t 1 t 1 t spatial description Observe that Ex, t ex, t and EX, t ex, t. t e y Y 1 t Yt t 1 Yt Yt 1 ex, t 1 0 1t 1 t 1t material description t e

33 .5 Variations of Distances Ch.. Deformation and Strain 33

34 Stretch The stretch ratio or stretch is defined as: stretch def PQ ds 0 PQ ds T t T t 0 Q t t Ω 0 X ds P dx x Ω dx P Q ds REMARK The sub-indexes ( ) T and ( ) t are often dropped. But one must bear in mind that stretch and unit elongation always have a particular direction associated to them. 34

35 Unit Elongation The extension or unit elongation is defined as: unit elongation def PQ dsds T t PQ ds T t 0 Q t t Ω 0 X ds P dx x Ω dx P Q ds REMARK The sub-indexes ( ) T and ( ) t are often dropped. But one must bear in mind that stretch and unit elongation always have a particular direction associated to them. 35

36 Relation between Stretch and Unit Elongation The stretch and unit elongation for a same point and direction are related through: ds ds ds ds ds If 1 0 ds ds : P and Q may have moved in time but have kept the distance between them constant. If 1 0 ds ds : the distance between them P and Q has increased with the deformation of the medium. If 1 0 ds ds : the distance between them P and Q has decreased with the deformation of the medium. 36

37 Stretch and Unit Elongation in terms of the Strain Tensors Considering: ds ds dxe dx ds ds dxe dx dx TdS dx t ds Then: ds ds ds ds ds ds 1 T E T 1 ds ds T E T 1 T E T ds ds t e t 1 t e t ds 1 ds 1 1 T E T t e t t e t REMARK E(X,t) and e(x,t) contain information regarding the stretch and unit elongation for any direction in the differential neighbourhood of a point. 37

38 .6 Variation of Angles Ch.. Deformation and Strain 38

39 Variation of Angles dx dx T ds T ds t 0 Ω 0 T (1) t Q t t () (1) T () Ω ds (1) R Q R Θ θ ds (1) X ds () ds () P P x dx dx t ds t ds The scalar product of the vectors dx (1) and dx () : dx dx dx dx cos ds ds cos 39

40 Variation of Angles d x 1 dx ds 1 ds cos 1 d x T d x d d ds ds ds ds 1 1 ds ds x x T E1 T T E1 T cos 1 ds ds 1 dx F dx dx F dx 1 1 T T dx dx FdX FdX dx F F dx E1 dx dx T ds T ds cos T 1 E T 1 T E T T E T 1 T E T T 1 E T T E T 1 T E T 40

41 Variation of Angles dx dx T ds T ds t 0 Ω 0 T (1) t Q t t () (1) T () Ω ds (1) R Q R Θ θ ds (1) X ds () ds () P P x dx dx t ds t ds The scalar product of the vectors dx (1) and dx () : dx dx dx dx cos ds ds cos 41

42 Variation of Angles d X 1 d X ds 1 ds 1 d X T cos d X 1 d X F dx 1 dx F dx 1 1 T T 1 dx dx F dx F dx dx F F dx 1e dx dx t ds t ds cos t 1 e t t e t 1 t e t dx dx ds t ( 1- e) t ds ds ds t ( 1- e) t ds ds cos 1 1 ds ds REMARK E(X,t) and e(x,t) contain information regarding the variation in angles between segments in the differential neighbourhood of a point. 1 1 t e t t 1 e t t e t 1 t e t 4

43 Example Let us consider the motion of a continuum body such that the spatial description of the Cartesian components of the spatial Almansi strain tensor is given by, tz 0 0 te ex (, t) te 0 t e e tz tz t Compute at time t=0, the length of the curve that at time t= is a straight line going from point a (0,0,0) to point b (1,1,1). The length of the curve at time t=0 can be expressed as, L B b 1 ds, t ds A x a ds 43

44 Example - Solution The inverse of the stretch, at the points belonging to the straight line going from a(0,0,0) to b(1,1,1) along the unit vector in the direction of the straight line, is given by, Where the unit vector is given by, 1 t x, t x, t 1 te x, t t 1 tex, t t Substituting the unit vector and spatial Almansi strain tensor into the expression of the inverse of the stretching yields, t x, t 1 te 3 1 T 44

45 Example - Solution The inverse of the stretch, which is uniform and therefore does not depends on the spatial coordinates, at time t= reads, 4 x, 1 e 3 1 Substituting the inverse of the stretch into the integral expression provides the length at time t=0, L 4 4 ds ds 3 e ds 3 e ds e b b (1,1,1) 1 x, a a (0,0,0) 3 45

46 .7 Physical Interpretation of E and e Ch.. Deformation and Strain 46

47 Physical Interpretation of E Consider the components of the material strain tensor, E: EXX EXY EXZ E11 E1 E13 E E E E E E E XY YY YZ 1 3 EXZ EYZ E ZZ E13 E3 E 33 For a segment parallel to the X-axis, the stretch is: reference configuration 1 T E T T E11 E1 E13 1 ET E E E 0 E T (1) T E13 E3 E33 0 (1) (1) T (1) T (1) dx dst ds 0 0 (1) 1E 1 11 Stretching of the material in the X-direction 47

48 Physical Interpretation of E Similarly, the stretching of the material in the Y-direction and the Z- direction: 1E 1 1E E 1 1E 1 1E 1 1E X X XX Y Y YY Z Z ZZ The longitudinal strains contain information on the stretch and unit elongation of the segments initially oriented in the X, Y and Z-directions (in the material configuration). E E E E E E E E E E XX XY XZ XY YY YZ XZ YZ ZZ If If E E XX YY X Y No elongation in the X-direction No elongation in the Y-direction If E ZZ 0 0 Z No elongation in the Z-direction 48

49 Physical Interpretation of E Consider the angle between a segment parallel to the X-axis and a segment parallel to the Y-axis, the angle is: 1 T 1 ET reference configuration T deformed configuration 1 cos 1 0 T 0 cos T E T 1 T E T E E 1 1 T T T E T E 11 1 T E T E 1 T E T E 1 E 11 EXY EXY xy arccos arcsin 1 E 1 E 1 E 1 E XX YY XX YY 49

50 Physical Interpretation of E EXY xy arcsin 1 E 1 E The increment of the final angle w.r.t. its initial value: XX YY reference configuration EXY XY xy arcsin XY 1 EXX 1 E YY deformed configuration 50

51 Physical Interpretation of E Similarly, the increment of the final angle w.r.t. its initial value for couples of segments oriented in the direction of the coordinate axes: The angular strains contain information on the variation of the angles between segments initially oriented in the X, Y and Z-directions (in the material configuration). E E E E E E E E E E XX XY XZ XY YY YZ XZ YZ ZZ XY XZ YZ arcsin EXY 1 EXX 1 E arcsin EXZ 1 EXX 1 E arcsin EYZ 1 E 1 E If If If EXY EXZ EYZ YY YY ZZ ZZ No angle variation between the X- and Y-directions No angle variation between the X- and Z-directions No angle variation between the Y- and Z-directions 51

52 Physical Interpretation of E In short, deformed configuration reference configuration dx 1E dx 1 3 XX dy 1E dy YY dz 1E dz ZZ XY XZ YZ EXY arcsin 1 E 1 E XX EXZ arcsin 1 E 1 E XX EYZ arcsin 1 E 1 E YY YY ZZ ZZ 5

53 Physical Interpretation of e Consider the components of the spatial strain tensor, e: exx exy e xz e11 e1 e13 e exy eyy e yz e1 e e 3 exz eyz e zz e13 e3 e 33 For a segment parallel to the x-axis, the stretch is: deformed configuration 1 1 t e t e11 e1 e13 1 tet e e e 0 e e13 e3 e 33 0 t (1) dx ds e 11 Stretching of the material in the x-direction 53

54 Physical Interpretation of e Similarly, the stretching of the material in the y-direction and the z- direction: x x 1 1 1e11 1exx 1 1 y y 1 1 1e 1e z z 1 1 1e 1e 33 The longitudinal strains contain information on the stretch and unit elongation of the segments oriented in the x, y and z-directions (in the deformed or actual configuration). yy zz e e e e e e e e e e xx xy xz xy yy yz xz yz zz 54

55 Physical Interpretation of e Consider the angle between a segment parallel to the x-axis and a segment parallel to the y-axis, the angle is: reference configuration deformed configuration t cos t cos t 1 e t t e t 1 t e t e e t t t e t e 11 1 t e t e 1 t e t e 1 e exy XY arcsin 1 e 1 xx e yy 55

56 Physical Interpretation of e exy XY arcsin 1 e 1 xx e yy The increment of the angle in the reference configuration w.r.t. its value in the deformed one: reference configuration exy xy xy XY arcsin 1 e 1 e deformed configuration xx yy 56

57 Physical Interpretation of e Similarly, the increment of the angle in the reference configuration w.r.t. its value in the deformed one for couples of segments oriented in the direction of the coordinate axes: exy xy XY arcsin 1 exx 1 e exz xz XZ arcsin 1 exx 1 e eyz yz YZ arcsin 1 e 1 e The angular strains contain information on the variation of the angles between segments oriented in the x, y and z-directions (in the deformed or actual configuration). e e e e e e e e e e xx xy xz xy yy yz xz yz zz yy yy zz zz 57

58 Physical Interpretation of e In short, reference configuration dx 3 1 dy dz 1e 1e 1e xx yy zz dx dy dz deformed configuration exy xy XY arcsin 1 e 1 e exz xz XZ arcsin 1 e 1 e eyz yz YZ arcsin 1 e 1 e xx xx yy yy zz zz 58

59 .8 Polar Decomposition Ch.. Deformation and Strain 59

60 Polar Decomposition Polar Decomposition Theorem: For any non-singular nd order tensor F there exist two unique positive-definite symmetrical nd order tensors U and V, and a unique orthogonal nd order tensor Q such that: not T left polar U F F decomposition not T V FF FQUVQ 1 1 QFU V F right polar decomposition The decomposition is unique. Q: Rotation tensor U: Right or material stretch tensor V: Left or spatial stretch tensor REMARK An orthogonal nd order tensor verifies: T T Q QQQ 1 60

61 Properties of an orthogonal tensor An orthogonal tensor Q when multiplied (dot product) times a vector rotates it (without changing its length): y Q x y has the same norm as x: T T y y T y yy Qx Qx xq Qx x 1 when Q is applied on two vectors x (1) and x (), with the same origin, the original angle they form is maintained: y y Qx Qx (1) (1) () () T T (1) () y y (1) () (1) T () (1) () y y x Q Qx x x cos (1) () (1) () (1) () y y y y x x Consequently, the rotation distances. y Q x maintains angles and 61

62 Polar Decomposition of F Consider the deformation gradient tensor, F: FQUVQ stretching rotation dxfdx VQ dxv QdX (not) F() stretching rotation() rotation stretching dxfdx QU dxq UdX not F() rotation stretching() REMARK For a rigid body motion: U V 1 and Q F 6

63 .9 Volume Variation Ch.. Deformation and Strain 63

64 Differential Volume Ratio Consider the variation of a differential volume associated to a particle P: reference configuration deformed configuration dv d d d 1 (3) 0 X X X dx1 dx dx 3 det dx1 dx dx 3 M dx1 dx dx 3 t M 1 x x 3 x dv d d d dx1 dx dx 3 det dx1 dx dx3 m dx1 dx dx3 m M dx m dx () i () i ij j ij j 64

65 Differential Volume Ratio Consider now: () i () i d d i{1,,3} x F X Fundamental eq. of deformation () i () i dx j Fjk dx k i, j {1,, 3} M dx and m dx () i () i ij j ij j m dx F dx F M M F mmf () i () i T T ij j jk k jk ik ik kj Then: dv t m M F M F F M F dv dv F dv0 T T 0 dv And, defining J (X,t) as the jacobian of the deformation, J X, t det F X, t 0 dvt J dv0 0 t 65

66 .10 Area Variation Ch.. Deformation and Strain 66

67 Surface Area Ratio Consider the variation of a differential area associated to a particle P: da: danmaterial vector "differential of area" da da da: danspatial vector "differential of area" da da Reference (initial) configuration Deformed (current) configuration 3 dv0 dh da d X NdA dh 3 3 dx NdAdAdX da 3 dvt dh da d x n da dh 3 3 dx nda dadx da 67

68 Surface Area Ratio Consider now: dv dadx t 3 3 (3) dx FdX dvt F dv0 dv dadx dv t F d AdX d afdx dx F dadaf F dv dv0 dvt 0 da F daf 1 da da nda NdA dan F NF 1 da da F NF 1 da 68

69 .11 Volumetric Strain Ch.. Deformation and Strain 69

70 Volumetric Strain Volumetric Strain: e X, t def X, X, X, dv t dv t dv dv dv t dv not 0 t 0 0 dvt F dv 0 e F dv dv 0 0 dv 0 e F 1 REMARK The incompressibility condition (null volumetric strain) takes the form e J 10 J F 1 70

71 .1 Infinitesimal Strain Ch.. Deformation and Strain 71

72 Infinitesimal Strain Theory The infinitesimal strain theory (also called small strain theory) is based on the simplifying hypotheses: t 0 P u Ω t P Displacements are very small w.r.t. the typical dimensions in the continuum medium, Ω 0 X x As a consequence, u ( size of 0) and the reference and deformed configurations are considered to be practically the same, as are the material and spatial coordinates: not x Xu X U X, t u X, t u x, t and 0 x X u X i i i i Displacement gradients are infinitesimal, i 1, i, j1,, 3. u x j not X X x U, t u, t u, t i{1,,3} i i i 7

73 Infinitesimal Strain Theory The material and spatial coordinates coincide, Even though it is considered that u cannot be neglected when calculating other properties such as the infinitesimal strain tensor ε. There is no difference between the material and spatial differential operators: The local an material time derivatives coincide x X u X 0 symb = eˆ ˆ i ei Xi xi JX (, t) UX (, t) ux (, t) j( x, t) ( X, t) ( x, t) ( x, t) ( X, t) x d X, t x, t dt t t 73

74 Strain Tensors 74 Green-Lagrange strain tensor 1 1 E F F1 JJ J J 1 u u i j uk u k Eij xj xi xi x i T T T Euler-Almansi strain tensor 1 1 e 1F F jj j j 1 u u i j uk u k eij xj xi xi x i T 1 T T Therefore, the infinitesimal strain tensor is defined as : not T T s JJ jj uu u 1 u u i j ij i, j{1,, 3} xj xi u x k j u x k j 1 1 T 1 T E JJ jj 1 u u i j Eij ij ij i, j{1,, 3} xj x i 1 T 1 T e jj JJ 1 u u i j eij ij i, j{1,, 3} xj x i REMARK ε is a symmetrical tensor and its components 1, i, j 1,, 3 are infinitesimal: ij

75 Stretch and Unit Elongation Stretch in terms of the strain tensors: T 1 T E T t x Considering that e E ε and that it is infinitesimal, a Taylor linear series expansion up to first order terms around x = 0 yields: 1 1 t e t x 1Tε T T 1x d 0 x1x dx x0 1 x 1 1 x d 0 x1x dx x0 1 1t ε t t But in Infinitesimal Strain Theory, T t. So the linearized stretch and unit elongation through a direction given by the unit vector T t are: ds ds ds 1t t 1T T 1 t t ds ds 75

76 Physical Interpretation of Infinitesimal Strains Consider the components of the infinitesimal strain tensor, ε: xx xy xz xy yy yz 1 3 xz yz zz For a segment parallel to the x-axis, the stretch and unit elongation are: reference configuration 1t t t ε t Stretch in the x-direction 1 x T t (1) (1) dx ds dx x xx Unit elongation in the x-direction

77 Physical Interpretation of Infinitesimal Strains Similarly, the stretching and unit elongation of the material in the y- direction and the z-direction: x x 1xx 1 y y 1 yy z z zz The diagonal components of the infinitesimal strain tensor are the unit elongations of the material when in the x, y and z-directions. xx xy xz xy yy yz xz yz zz 77

78 Physical Interpretation of Infinitesimal Strains Consider the angle between a segment parallel to the X-axis and a segment parallel to the Y-axis, the angle is XY. Applying: EXY xy arcsin 1 E 1 E XX YY E E E XX XY YY xx xy yy reference configuration xy arcsin arcsin 1 1 xx yy xy 1 1 xy xy xy REMARK The Taylor linear series expansion of arcsin x yields d arcsin arcsin x arcsin 0 x... xo x dx x0 78

79 Physical Interpretation of Infinitesimal Strains xy xy The increment of the final angle w.r.t. its initial value: xy xy xy xy Similarly, the increment of the final angle w.r.t. its initial value for couples of segments oriented in the direction of the coordinate axes: ; ; xy xy xz xz yz yz The non-diagonal components of the infinitesimal strain tensor are equal to the semi-decrements produced by the deformation of the angles between segments initially oriented in the x, y and z-directions. 79

80 Physical Interpretation of Infinitesimal Strains In short, reference configuration deformed configuration xx yy zz x z y xy xz yz xy xz yz 80

81 Engineering Strains Using an engineering notation, instead of the scientific notation, the components of the infinitesimal strain tensor are REMARK Positive longitudinal strains indicate increase in segment length. Angular strains Longitudinal strains Positive angular strains indicate the corresponding angles decrease with the deformation process. 81 Because of the symmetry of ε, the tensor can be written as a 6-component infinitesimal strain vector, (Voigt s notation): def 6 R x y z xy xz yz [,,,,, ] longitudinal strains angular strains T

82 Variation of Angles Consider two segments in the reference configuration with the same origin an angle Θ between them. cos T 1 E T T E T 1 T E T reference configuration cos( ) E ε T T (1) () 1T T 1T T deformed configuration 1 1 (1) (1) () () cos( ) T T T T (1) () (1) () 8

83 Variation of Angles cos( ) T T T T (1) () (1) () T (1) and T () are unit vectors in the directions of the original segments, (1) () (1) () therefore, T T T T cos cos T t Also, cos( ) cos cos sinsin cos sin 1 cos sin cos T T (1) () (1) (1) T t () () T T t t sin sin (1) () (1) () REMARK The Taylor linear series expansion of sin x and cos x yield sin x sin 0 d sin dx x0 x... xo x cos x cos 0 d cos dx x... 1O x x0 83

84 Polar Decomposition Polar decomposition in finite-strain problems: right polar decomposition not T U F F not T V FF FQUVQ 1 1 QFU V F left polar decomposition REMARK In Infinitesimal Strain Theory x X, therefore, x F X 1 84

85 Polar Decomposition In Infinitesimal Strain Theory: 1 U F F J J JJ J J JJ JJ J x T T T T T T U 1 infinitesimal strain tensor 85 Similarly, U J J T x 1 1 U 1 1 infinitesimal strain tensor REMARK The Taylor linear series expansion of 1 x and 1 x 1 yield d 1 x 1x 0 x1 xo x dx d dx x0 x1x 1 0 x1xox x0

86 Polar Decomposition T T T T QFU J JJ J J J J J J J J J Q 1 The infinitesimal rotation tensor Ω is defined: def def 1 T 1 a ( JJ ( uu) u 1 u u i j ij 1 i, j {1,,3} xj xi The diagonal terms of Ω are zero: It can be expressed as an infinitesimal rotation vector θ, def u1 u 3 31 x3 x1 1 u u u3 u x x 1 1 u x x 0 REMARK The antisymmetric or skew-symmetrical gradient operator is defined as: 1 a () () () REMARK Ω is a skew-symmetric tensor and its components are infinitesimal. 86

87 Polar Decomposition From any skew-symmetric tensor Ω, it can be extracted a vector θ (axial vector of Ω) exhibiting the following property: r r r As a consequence: The resulting vector is orthogonal to r. If the components of Ω are infinitesimal, then r r is also infinitesimal The vector r r r r can be seen as the result of applying a (infinitesimal) rotation (of axial vector θ) on the vector r. 87

88 Proof of θr Ωr r The result of the dot product of the infinitesimal rotation tensor, Ω, and a generic vector, r, is exactly the same as the result of the cross product of the infinitesimal rotation vector, θ, and this same vector r r r r r r 3 Proof: r eˆ1 eˆ eˆ3 eˆ ˆ ˆ 1 e e3 1r 31r3 not r r r1 r r3 r1 r r 3 31r1 3r r r1 1r 31r3 0 r r r r 3 31r1 3r 88

89 Polar Decomposition Using: J F1 1 JJ Q 1 T 1 1 F J JJ JJ T T 1 1 F 1 Consider a differential segment dx: stretch rotation dxfdx1dx dx 1 dx F() stretching () rotation() 89 REMARK The infinitesimal rotation tensor characterizes the rotation and, in the small-strain context, maintains angles and distances.

90 Volumetric Deformation The volumetric strain: e F 1 Considering: F Q Uand U 1 1 F QU Q U U 1 det 1 1xx yy zz O 1Tr Tr xx xy xz xy yy yz 1 xz yz zz e Tr 90

91 .13 Strain Rate Ch.. Deformation and Strain 91

92 Spatial Velocity Gradient Tensor Consider the relative velocity between two points in space at a given (current) instant: vp v( x, t) v( x1, x, x3, t) dvx (, t) vq vp v xdx, t v x, t v dv dxldx x l vi dvi dxj lij dxj x j l i, j1,,3 ij Spatial velocity gradient tensor def v x, t lx, t v x vi lij i, j1,,3 x j 9

93 Strain Rate and Rotation Rate (or Spin) Tensors The spatial velocity gradient tensor can be split into a symmetrical and a skew-symmetrical tensor: l v vi lij i, j 1,,3 x j l sym l skew l : dw def 1 1 d sym() v v v 1 v v i j d ij i, j{1,,3} xj xi d11 d1 d31 d1 d d 3 d31 d3 d 33 d Strain Rate Tensor Rotation Rate or Spin Tensor not T s def l l l 1 T 1 skew () l l l a w v v v 1 v v i j w ij i, j{1,,3} xj xi 0 w1 w31 w w1 0 w 3 w31 w3 0 not 93

94 Physical Interpretation of d The strain rate measures the rate of deformation of the square of the differential length ds in the spatial configuration, d d d d d d ds() t d d d d d d d x d d d x x x x x x x x x dvdxdxdv dt dt dt dt dt dt dv ldx v v 1 ( T d ll ) d T T ds () t d x d x d x d x d x d d d dt x x d x l l l l T d dv dv Differentiating w.r.t. time the expression ds() t ds dxedx d d de d ( ( ds()) t ( ds) ) d, td d d ( ds()) t dt dt dt dt constant dxe X, t dx XEX X X X notation E 94

95 Physical Interpretation of d dxe dxdxd dx dx FdX And, rearranging terms: T T T ( ) dxe dx dxddx dx d dx FdX d FdX dx F df dx FdX T d X F d F E d X 0 d X F d X F X T T d d X F T F dfe 0 There is a direct relation between the material derivative of the material strain tensor and the strain rate tensor but they are not the same. E and d will coincide when in the REMARK reference configuration F tt0 1. Given a nd order tensor A, if xax 0for any vector x 0 then A 0. T E F df 95

96 Physical Interpretation of w To determine the (skew-symmetric) rotation rate (spin) tensor only three different components are needed: The spin vector (axial vector [w]) of can be extracted: The vector is named vorticity vector w w w w w w w v v v v ( ) v v x x w rot w x x w x x v v v v v 1 w, {1,,3} j i ij j i i j x x w 96

97 Physical Interpretation of w It can be proven that the equality r wr r holds true. Therefore: ω is the angular velocity of a rotation movement. ω x r = w r is the rotation velocity of the point that has r as its position vector w.r.t. the rotation centre. Consider now the relative velocity dv, dv l dx l dw dvddxwdx 97

98 Proof of r wr r The result of the dot product of the rotation rate tensor, w, and a generic vector, r, is exactly the same as the result of the cross product of the axial rotation rate vector, ω, and this same vector. r wr r Proof: eˆ1 eˆ eˆ3 eˆ1 eˆ eˆ3 wr 13 3 wr 1 not r 1 3 w3 w13 w 1 w1r1w3r3 r r r r r r w r w r w1 w13 r1 w1r w13r3 w 0 w wr r w r w r w13 w3 0 r 3 w13rw3r 1 98

99 .14 Material time Derivatives Ch.. Deformation and Strain 99

100 Deformation Gradient Tensor F The material time derivative of the deformation gradient tensor, xi ( X, t) Fij ( X, t) i, j{1,,3} X j d dt REMARK The equality of cross derivatives applies here: () () i j j i X X X v i xx, t dfij xi, t xi, t Vi, t x k l dt t X j X j t X j xk X j = V( X, t) = l ik = F kj notation df F lf dt dfij F ij lik Fkj i, j1,,3 dt i ik F kj 100

101 Inverse Deformation Gradient Tensor F -1 The material time derivative of the inverse deformation gradient tensor, Rearranging terms, d F dt 1 1 FF 1 d dt 1 d F d 1 df 1 ( FF ) F F 0 dt dt dt 1 d F df 1 1 F F F F dt dt l l l F F FF F FF F REMARK Do not mistake the material derivative of the inverse tensor for the inverse of the material derivative of the tensor: d dt 1 F X, t F X, t d F 1 1 F dt 1 dfij dt F 1 l 1 ik kj l i, j 1,,3 101

102 Strain Tensor E The material time derivative of the material strain tensor has already been derived for the physical interpretation of the deformation rate tensor: T E F df A more direct procedure yields the same result: 1 T E F F1 d dt F lf F F l T T T de 1 T T 1 T T T T 1 T T dt E F F F F F l F F l F F l l F F d F d T E F df 10

103 Strain Tensor e The material time derivative of the spatial strain tensor, 1 T e 1 F F d dt 1 F F F l 1 1 l F T T T de dt 1 1 e F F F F l F F F F l T 1 T 1 T T 1 T 1 1 T T 1 T 1 e l F F F F l 103

104 Volume differential dv The material time derivative of the volume differential associated to a given particle, 0 dv xx (, t), t FX (, t) dv ( X) d dt d FX (, t) d dv t dv dv dt t dt 0 F 0 d dv v F dv 0 dt dv d dv ( x, t ) v( x, t ) dv ( x, t ) dt The material time derivative of the determinant of the deformation gradient tensor is: For a nd order tensor A: d F d F df df F dt df dt dt ij 1 ij 1 Fji FkjFji lik ij 1 l F FF ik vi d F F l ii F F v F v ( v) F xi dt v d A d A d A A A daij ij kj F ki 1 ji ki 104

105 Area differential vector da The material time derivative of the area differential associated to a given particle, 1 1 (, ), (, ) ( ) (, ) da x X t t F X t da X F X t F da F d dt d d F d da t da F F da F dt dt dt F v 1 1 F 1 l d d 1 1 d d a v F AF F AF l dt da da d d d d d d d dt a a v al a1( v) al a( v) 1l da1 105

106 .15 Other Coordinate Systems Ch.. Deformation and Strain 106

107 Coordinate Systems The equations expressed in intrinsic or compact notation are independent of the coordinate system used. The components of the vectors and tensors do depend on the coordinate system used. The most commonly used is the Cartesian coordinate system. Depending on the problem being dealt with, other systems are more convenient: Cylindrical coordinate system Spherical coordinate system 107

108 Curvilinear Orthogonal Coord. System A curvilinear coordinate system is defined by: The coordinates, generically named Its vector basis, eˆ ˆ ˆ a, eb, ec, formed by unit vectors eˆ eˆ eˆ 1. If the elements of the basis are orthogonal is is called an orthogonal coordinate system: eˆ eˆ eˆ eˆ eˆ eˆ 0 The orientation of the curvilinear basis may change at each point in space,. eˆ ˆ m em( x) m{ a, b, c} abc,, a b a c b c a b c REMARK A curvilinear orthogonal coordinate system can be seen as a mobile Cartesian coordinate system x, y, z, associated to a curvilinear basis eˆ, eˆ, eˆ. a b c 108

109 Curvilinear Orthogonal Coord. System A curvilinear orthogonal coordinate system can be seen as a mobile Cartesian coordinate system eˆ ˆ ˆ a, eb, ec, associated to a curvilinear basis x, y, z. The components of a vector and a tensor magnitude in the curvilinear orthogonal basis will correspond to those in the given Cartesian local system: v v T T T T T T v a x aa ab ac xx xy xz vb v y Tba Tbb T T bc Tyx Tyy Tyz v c v z Tca Tcb Tcc Tzx Tzy T zz The components of the curvilinear operators will not be the same as those in the given Cartesian local system. They must be obtained for each specific case. 109

110 Cylindrical Coordinate System z coordinate line coordinate line r coordinate line x r cos x(, r, z) y r sin z z eˆ r eˆ eˆ eˆ r dv r d dr dz 110

111 Cylindrical Coordinate System Nabla operator r 1 1 eˆ ˆ ˆ r e e z r r z r z Displacement vector ur uu ˆ u ˆ u ˆ rer e zez u u u z x r cos x(, r, z) y r sin z z Velocity vector vr v v ˆ v ˆ v ˆ rer e zez u v v z 111

112 Cylindrical Coordinate System Infinitesimal strain tensor rr zz x x xy xz rr r rz 1 T u u x y yy yz r z x z yz zz rz z zz u r r 1 u u r r u z z r r rz z 11 u r u u r r r 1 ur u z z r 1u 1 u z z r x r cos x(, r, z) y r sin z z 11

113 Cylindrical Coordinate System Strain rate tensor dx x dxy dxz drr dr drz 1 T d v v dx y dyy d yz dr d d z dx z dyz d zz drz d z d zz d d d rr zz vr r 1 v r v z z v r r d d d r rz z 11 vr v v r r r 1 vr v z z r 1v 1 v z z r x r cos x(, r, z) y r sin z z 113

114 Spherical Coordinate System x rsin cos xxr,, y r sin sin z r cos r coordinate line eˆ ˆ ˆ r e e eˆ ˆ er 0 dv r sin dr d d 114

115 Spherical Coordinate System r eˆ ˆ ˆ r e e r r rsin r 1 r sin Nabla operator Displacement vector u r uu ˆ u ˆ u ˆ rer e e u u u x rsin cos xxr,, y r sin sin z r cos Velocity vector vr v v ˆ v ˆ v ˆ rer e e u v v 115

116 Spherical Coordinate System rr Infinitesimal strain tensor xx xy xz rr r r 1 T u u xy yy yz r xz yz zz r u r r 1 u ur r r x rsin cos 1 u u u xxr,, y r sin sin r cotg z r cos r sin r r r r 11 u r u u r r r 1 1 u u u r rsin r r 1 1 u 1 u u rsin r r cotg 116

117 Spherical Coordinate System Deformation rate tensor d d d d d d rr r r dxx dxy dxz drr dr d r 1 T d v v dxy dyy dyz dr d d dxz dyz d zz dr d d vr r x rsin cos 1 v v xx r r,, y r sin sin r r z r cos 1 v v vr cotg r sin r r 11 vr v v r r r 1 1 v v v r rsin r r 1 1 v 1 v v rsin r r cotg 117

118 Summary Ch.. Deformation and Strain 118

119 Summary (Material) deformation gradient tensor: The inverse deformation gradient tensor or spatial deformation gradient tensor: not 1 F x, t X Properties: not 1 ij X x 1 1 FF F F1 If F Ft, the deformation is said to be homogeneous. If there is no motion: 1 FF 1 and xx F j i i, j 1,,3 F X, t F ij not not xi X x j i, j 1,,3 119

120 Summary (cont d) Displacement field Material description: (Lagrangian form), t, t X, X, 1,,3 U X x X X Ui t xi t Xi i Displacement Gradient Tensor Material description: Spatial description: (Eulerian form) ux, t xxx, t u x, tx X x, t i1,,3 i i i Spatial description: def, t, t J X U X F1 Ui Jij Fij ij i, j1,,3 X Properties: j, t, t 1 If there is no motion, F F 1 and J = j = 0. def jx ux 1 F i 1 ij ij ij x j 1 u j F i, j 1,,3 10

121 Summary (cont d) Strain is a normalized measure of deformation which represents the displacement between particles in the body relative to a reference length. Strain Tensor Material description: (Green-Lagrange Strain Tensor) Properties of the Strain Tensors Both are symmetrical They are not the material and spatial descriptions if a same strain tensor. They are affected by different vectors (dx and dx): ds ds dxe dxdxe dx Spatial description: (Euler-Almansi Strain Tensor) 1 T 1 T, t 1 ex, t 1 F F E X F F 1 Eij t FkiFkj ij i j X,, 1,,3 1 1 eij t ij Fki Fkj i j 1 1 x,, 1,,3 11

122 Summary (cont d) Strain is a normalized measure of deformation which represents the displacement between particles in the body relative to a reference length. Strain Tensor Material description: (Green-Lagrange Strain Tensor) Properties of the Strain Tensors Both are symmetrical Spatial description: (Euler-Almansi Strain Tensor) 1 T 1 T, t 1 ex, t 1 F F E X F F 1 Eij t FkiFkj ij i j X,, 1,,3 1 T 1 T T E J J JJ J J 1 U U i j Uk U k Eij i, j{1,, 3} X j Xi Xi X j 1 1 eij t ij Fki Fkj i j 1 1 x,, 1,,3 1 T 1 T T e j j jj j j 1 u u i j uk u k eij i, j1,,3 xj xi xi xj They are not the material and spatial descriptions if a same strain tensor. They are affected by different vectors (dx and dx): dxe dxdxe dx 1

123 Summary (cont d) Stretch ratio or stretch: Extension or elongation ratio or unit elongation: PQ ds (0 ) PQ ds T t PQ ds ds T t PQ ds In terms of strain tensors: 1 T E T 1 1 T E T1 1 1 t e t t e t ds dx dx ds dx dx 13

124 Summary (cont d) Variation of angles: cos 1 T E T T E T 1 T E T cos t 1 e t t e t 1 t e t dx dx T ds t ds dx dx T ds t ds 14

125 Summary (cont d) Physical Interpretation of E deformed configuration E E E E E E E E E E XX XY XZ XY YY YZ XZ YZ ZZ reference configuration dx 1E dx 1 dy 1E dy dz 1E dz 3 XX YY ZZ XY XZ YZ EXY arcsin 1 E 1 E XX EXZ arcsin 1 E 1 E XX EYZ arcsin 1 E 1 E YY YY ZZ ZZ 15

126 Summary (cont d) Physical Interpretation of e reference configuration dx 1 dy 1 1 e 1 xx 1 e yy dx dy deformed configuration dz e zz dz exy xy XY arcsin 1 e 1 e exz xz XZ arcsin 1 e 1 e eyz yz YZ arcsin 1 e 1 e xx xx yy yy zz zz e e e e e e e e e e xx xy xz xy yy yz xz yz zz 16

127 Summary (cont d) Polar Decomposition: not T U F F not T V FF FQUVQ 1 1 QFU V F left polar decomposition The decomposition is unique. Q: Rotation tensor U: Right or material stretch tensor V: Left or spatial stretch tensor Polar decomposition of F : right polar decomposition not F() deformation rotation() not F() rotation deformation() 17

128 Summary (cont d) Differential Volume Ratio dvt F dv 0 dvt J dv 0 J (X,t) is the jacobian of the deformation, X FX J, t det, t 0 Surface Area Ratio d da NdA da nda 1 1 a F daf da F NF da 18

129 Summary (cont d) Volumetric Strain e X, t def X, X, X, dv t dv t not dvt dv0 dv t dv 0 e F 1 Spatial velocity gradient tensor def v x, t lx, t x l v l ij vi x j i, j 1,,3 l sym l skew l : dw 19

130 Summary (cont d) def 1 1 d sym() v v v 1 v v i j d ij i, j{1,,3} xj xi d d d not def T s T l l l skew () l l l d d1 d d 3 d31 d3 d 33 Strain Rate Tensor T E F df 1 1 not a w v v v 1 v v i j w ij i, j{1,,3} xj xi 0 w1 w w w 0 w w31 w Rotation Rate or Spin Tensor Axial (dual) rotation rate vector w rot( v) v w 13 w 1 3 r wr r 130 dvddxwdx

131 Summary (cont d) Material derivatives of the deformation gradient tensor: of the inverse deformation gradient tensor : of the material strain tensor : of the spatial strain tensor : T E F df of the deformation gradient tensor : not df F lf dt dfij F ij lik Fkj i, j1,,3 dt of the volume differential associated to a given particle: of the area differential associated to a given particle: 1 d F dt df dt 1 1 F 1 ij T T 1 T 1 e l F F F F l d F dt F v F l 1 ik kj l i, j 1,,3 d dv v dv dt d d a d a v1 l dt 131

132 Summary (cont d) Infinitesimal Strain Theory (small deformation theory) u ui x j X 0 1, i, j 1,, 3 and x Xu X x X u X i i i i not, t, t, t U X u X u x not X X x U, t u, t u, t i{1,,3} i i i Strain tensors in material and spatial descriptions coincide. An Infinitesimal Strain Tensor is defined: not Stretch: Unit Elongation: 1 T s Ee JJ u 1 u u i j Eij eij ij i, j{1,,3} xj xi 1t t 1T T 1 t t 13

133 Summary (cont d) Physical Interpretation of infinitesimal strains: reference configuration unit elongations deformed configuration xx x yy zz y z xy xz yz xy xz yz 133 engineering notation 1 1 scientific notation x xy xz xx xy xz 1 1 xy yy yz xy y yz xz yz zz 1 1 xz yz z Angular strains R def 6 x, y, z, xy, xz, yz Longitudinal strains infinitesimal strain vector: T

134 Summary (cont d) Infinitesimal Strain Theory Variation of angles: Polar decomposition: U 1 Q 1 Infinitesimal Rotation Tensor: T T t t sin sin F (1) () (1) () def def 1 T 1 a ( JJ ( uu) u 1 u u i j ij 1 i, j {1,,3} xj xi 1 F() deformation() rotation() 134 Infinitesimal Rotation Vector: 3 def 1 31 u 1 r r r

135 Summary (cont d) Infinitesimal Strain Theory Volumetric Deformation: e Tr Coordinate Systems The equations expressed in intrinsic or compact notation are independent of the coordinate system used. The components of the vectors and tensors do depend on the coordinate system used. As will the components of the operators. The Cartesian coordinate system is the most commonly used. Other systems are: Cylindrical coordinate system Spherical coordinate system x r cos x(, r, z) y r sin z z x r sin cos xxr,, y r sin sin z r cos 135